Follow the oxygen atom, the carbon bearing OH and the carbon skeleton through preparation, oxidation, substitution and elimination.
JEE Main · JEE Advanced · NEET UG48 min study guide10 indexed sections
BY THE END, YOU CAN
Classify alcohols by the carbon directly bearing OH, including allylic and benzylic cases.
Choose carbonyl reduction, Grignard addition or alkene hydration from a target structure.
Distinguish primary, secondary and tertiary oxidation outcomes and reagent scope.
Explain activation of OH before substitution and the stereochemical consequences of the pathway.
Predict dehydration products with rearrangement and competition caveats.
Track atoms through esterification, ether formation, diol cleavage and protecting-group strategies.
BEFORE YOU START
Nucleophilic substitution, elimination and carbonyl addition.
Alcohol acidity, alkoxide formation and stereochemical vocabulary.
Carbon counting and oxidation-state changes.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. Classify the reacting centre rather than the whole molecule
An alcohol has an OH group attached to a saturated carbon. Phenols have OH directly attached to an aromatic ring, and enols have OH on an alkene carbon; their reactions require separate reasoning. Benzyl alcohol, C₆H₅CH₂OH, is an alcohol because the OH-bearing carbon is the side-chain CH₂, whereas C₆H₅OH is phenol.
A primary alcohol has an OH-bearing carbon attached to one other carbon, a secondary alcohol to two, and a tertiary alcohol to three. Methanol is a special zero-carbon-neighbour case often discussed with primary-alcohol chemistry. Count neighbours of the correct carbon: branching elsewhere does not turn a primary alcohol into a tertiary one.
An allylic alcohol has its OH-bearing saturated carbon adjacent to a C=C bond; a benzylic alcohol has it adjacent to an aromatic ring. These labels describe position and can coexist with primary or secondary classification. Resonance stabilization of a possible intermediate may make an allylic or benzylic substrate behave differently from an ordinary unactivated alcohol.
Two different bonds can react. Acid–base chemistry usually breaks O–H, leaving the C–O bond intact. Conversion to an alkyl halide breaks the C–O connection at the reacting carbon. Oxidation changes bonding at that carbon while generally retaining the oxygen in the simple introductory picture. Esterification typically changes bonding at oxygen without replacing the entire carbon skeleton.
Draw the OH-bearing carbon and all of its immediate substituents before choosing a reaction. That small local diagram determines whether oxidation can form a carbonyl without C–C cleavage, whether an SN2 approach is sterically plausible and whether the product introduces a new stereocentre.
02 / UNDERSTAND & APPLY
2. Reduction prepares alcohols without adding carbon atoms
Reduction of an aldehyde RCHO gives RCH₂OH, a primary alcohol. Reduction of a ketone R₂CO gives R₂CHOH, a secondary alcohol. In each case the carbonyl carbon remains in the skeleton and gains a hydrogen. A simple carbonyl hydride reduction cannot create a tertiary alcohol because the newly added group is H, not an additional carbon substituent.
In electron-flow terms, a hydride equivalent is delivered to the electrophilic carbonyl carbon while the C=O pi pair moves to oxygen. The intermediate is an alkoxide. Protonation in the subsequent appropriate workup produces the neutral alcohol. Drawing a carbonyl carbon with an extra bond but leaving its pi bond intact would violate its octet.
NaBH₄ commonly reduces aldehydes and ketones under standard teaching conditions while leaving an ordinary ester or carboxylic acid unreduced. LiAlH₄ is more broadly reducing and can convert esters and carboxylic acids to alcohols as well. Reagent names represent different chemoselectivity, not merely different spellings of “add hydrogen”. Specialized conditions can change reactivity, so broad rules should be tied to ordinary specified conditions.
An ester RCOOR′ gives RCH₂OH from its acyl portion and R′OH from its alkoxy portion after complete hydride reduction and workup. The ester’s two carbon fragments do not automatically join into one longer alcohol. An acid RCO₂H gives RCH₂OH with its carbonyl carbon retained. A pre-existing acidic OH also consumes reducing reagent through acid–base chemistry.
An achiral planar ketone reduced with an achiral reagent in an achiral environment often gives equal enantiomers if a new stereocentre is formed and the two faces are otherwise equivalent. Existing stereocentres or chiral catalysts can make the faces nonequivalent. Do not label every ketone reduction product racemic without checking whether the product is even chiral.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Formal hydride-equivalent electron bookkeeping, with reagent coordination omitted. Real hydride is delivered from a reagent rather than a freely dissolved hydride ion. Intermediate bonds are drawn flat for connectivity, not as a planar molecular geometry.
ORIGINAL WORKED EXAMPLE
Reduce both parts of an ester
Predict the alcohol products from complete LiAlH₄ reduction and workup of propyl ethanoate, CH₃COOCH₂CH₂CH₃.
Separate the acyl fragment CH₃CO– from the alkoxy fragment –OCH₂CH₂CH₃.
The acyl carbon becomes CH₂OH, giving CH₃CH₂OH.
The propoxy fragment becomes HOCH₂CH₂CH₃.
The total carbon count is 2+3=5, matching the ester.
Answer Ethanol and propan-1-ol.
03 / UNDERSTAND & APPLY
3. Grignard addition adds a new carbon substituent
An organomagnesium reagent RMgX behaves as a strongly basic carbon nucleophile in the standard carbonyl-addition model. Its R group bonds to the carbonyl carbon, the pi pair moves onto oxygen and the resulting alkoxide is protonated during workup. The C–C bond formation and protonation are separate conceptual steps.
Addition to formaldehyde followed by workup gives RCH₂OH, extending the reagent’s carbon skeleton by one carbon. Addition to an ordinary aldehyde R′CHO gives a secondary alcohol R′CH(OH)R. Addition to a ketone R′COR″ gives a tertiary alcohol R′C(OH)(R)R″. These classifications follow directly from counting carbon neighbours at the former carbonyl carbon.
An ordinary ester commonly undergoes two additions of the same Grignard R group at its acyl carbon, via an intermediate ketone, to give a tertiary alcohol with two identical newly introduced groups. The alkoxy portion leaves and becomes its corresponding alcohol on workup. Formate esters are an important structural exception: the retained substituent is H, so two additions give a secondary alcohol rather than a tertiary one.
Water, alcohols, carboxylic acids and other sufficiently acidic X–H groups quench a Grignard reagent by proton transfer. Thus adding RMgX to an unprotected hydroxy carbonyl compound is not described by carbonyl addition alone. The functional-group inventory determines reagent consumption and may require a protecting-group strategy.
Retrosynthesis works backward from the OH-bearing carbon. For a tertiary target, disconnect one C–C bond to assign one substituent to the Grignard reagent and leave the other two on a ketone. Equivalent substituents reduce the number of distinct disconnections. Carbon accounting is more reliable than memorizing a list of target names.
ORIGINAL WORKED EXAMPLE
Design a tertiary alcohol from a ketone
Give two distinct ketone/Grignard disconnections for 3-methylhexan-3-ol, whose OH-bearing carbon carries methyl, ethyl and propyl groups.
Disconnect the methyl group: the ketone retains ethyl and propyl groups, so use hexan-3-one with methylmagnesium halide, then workup.
Alternatively disconnect ethyl: the ketone retains methyl and propyl groups, so use pentan-2-one with ethylmagnesium halide, then workup.
Each route gives the same central carbon attached to OH, CH₃, CH₂CH₃ and CH₂CH₂CH₃.
The target has seven carbons, consistent with 6+1 or 5+2.
Answer Hexan-3-one + MeMgX, or pentan-2-one + EtMgX, each followed by appropriate workup.
04 / UNDERSTAND & APPLY
4. Select hydration by regioselectivity and rearrangement risk
Acid-catalysed hydration of an alkene adds H and OH overall. In the usual carbocation pathway, protonation forms the more favourable cation and water captures it, giving a Markovnikov orientation for simple unsymmetrical alkenes. Carbocation rearrangements can alter the skeleton or OH position, so the initial alkene substitution pattern is not always the final answer.
Oxymercuration followed by reduction gives a common teaching route to Markovnikov alcohols without a freely rearranging carbocation. Its bridged intermediate explains why ordinary carbocation shifts are avoided. The overall transformation should not be assigned a universal strict syn or anti stereochemistry simply from one intermediate step, because the reduction stage complicates the stereochemical outcome.
Hydroboration followed by oxidation places OH at the less substituted alkene carbon in common simple examples, with overall syn addition of H and OH and no carbocation rearrangement. The OH orientation follows replacement of C–B by C–O during oxidation. For a terminal alkene, this often provides the corresponding primary alcohol.
An alkyl halide can also be converted to an alcohol by substitution with a suitable aqueous nucleophile. A primary substrate may react by SN2; tertiary or resonance-stabilized substrates can follow ionization pathways under appropriate conditions. Elimination competes, especially with hindered substrates and basic conditions. “Aqueous hydroxide gives alcohol” is therefore a useful pattern, not a guarantee for every substrate.
These routes create different bonds. Hydration uses an existing C=C and retains carbon count. Carbonyl reduction changes C=O without adding carbon. Grignard addition forms a new C–C bond. Choosing a route is often easiest when the target’s carbon skeleton is compared with the starting material before considering reagents.
ORIGINAL WORKED EXAMPLE
Choose between two regioisomeric alcohols
Starting from hex-1-ene, choose standard conceptual hydration routes to hexan-1-ol and hexan-2-ol.
To put OH on terminal C1, use hydroboration–oxidation: H/OH add with the usual anti-Markovnikov orientation.
To put OH on C2 without a free-carbocation rearrangement pathway, use oxymercuration–reduction.
Both retain the six-carbon chain. Neither route adds a carbon atom.
5. Oxidation depends on a hydrogen at the OH-bearing carbon
A primary alcohol can be oxidized to an aldehyde and, with suitable further oxidizing conditions, to a carboxylic acid. A secondary alcohol gives a ketone. A tertiary alcohol lacks a hydrogen on its OH-bearing carbon and cannot form a carbonyl there by the ordinary two-hydrogen removal without breaking a C–C bond. Harsh oxidative cleavage is a different process, so “tertiary alcohols never react with any oxidant” is too absolute.
For a primary substrate, stopping at the aldehyde requires an appropriate reagent and conditions that limit further oxidation. PCC is a standard textbook example under its usual nonaqueous conditions; other selective oxidation systems exist. Aqueous strongly oxidizing conditions commonly take primary alcohols to acids. Reagent identity and medium determine the endpoint.
Aldehydes can hydrate in water to geminal diols, which helps explain why further oxidation can occur under suitable aqueous conditions. Ketones do not generally undergo the same simple oxidation to a same-carbon-count acid. The distinction is structural rather than a vague statement that one product is “more reactive”.
For an ordinary primary alcohol RCH₂OH, the OH-bearing carbon’s oxidation state changes from−1 to+1 in RCHO when R begins with carbon: a two-electron oxidation. Further oxidation to RCO₂H takes it to+3, another two-electron change. A secondary alcohol R₂CHOH has that carbon at 0 and a ketone at+2. Methanol’s carbon starts at−2, so use actual attached atoms rather than forcing it into the R=carbon case.
Oxidation of an enantiopure secondary alcohol to an achiral ketone can erase the stereocentre at the reacting carbon. Subsequent reduction with an achiral reagent can then regenerate both enantiomers, assuming no other stereochemical influence. Oxidation is therefore not merely a formula change; it can change stereochemical information.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Connectivity map, not a complete reagent mechanism. The aldehyde endpoint requires suitable selective conditions; further oxidation uses compatible conditions. Tertiary alcohols can undergo other harsh oxidative cleavage chemistry, which is not shown.
ORIGINAL WORKED EXAMPLE
Choose an oxidation endpoint
Predict the standard products of oxidizing pentan-1-ol under aldehyde-selective conditions and under suitable aqueous conditions allowing complete oxidation, and compare pentan-2-ol.
Pentan-1-ol is primary, so selective two-electron oxidation gives pentanal.
Further oxidation at the same terminal carbon gives pentanoic acid.
Pentan-2-ol is secondary and gives pentan-2-one in ordinary alcohol oxidation.
No carbon atoms are added or removed in these specified transformations.
Answer Pentanal; pentanoic acid; and pentan-2-one, respectively.
06 / UNDERSTAND & APPLY
6. Activate OH before asking it to leave
Unprotonated hydroxide is a poor leaving group from an ordinary alcohol in substitution. Protonation converts OH into an attached water group, which can leave more readily. Alternatively, reaction at oxygen can install a sulfonate leaving group or another activated derivative while initially retaining the C–O bond.
With hydrogen halides, a tertiary alcohol can ionize after protonation and then be captured by halide, often through SN1-type chemistry. Rearrangement and elimination must be considered where relevant. Primary alcohols generally avoid an unstable free primary carbocation and can undergo direct displacement of the activated group under suitable conditions.
PBr₃ commonly converts suitable primary and secondary alcohols to bromides through oxygen activation followed by backside displacement at carbon. At a stereogenic carbon, the SN2 step produces inversion of geometry. The R/S letter is not itself a reliable synonym for inversion because substituent priorities can change when OH is replaced by Br.
SOCl₂ can convert alcohols to chlorides, but stereochemical outcome depends on solvent, additives and the pathway. With pyridine under standard SN2-promoting conditions, inversion is often taught; other conditions can give retention through an internal-return pathway. Do not write “SOCl₂ always retains” or “always inverts” without the stated conditions.
Tosylation or mesylation takes place at oxygen and does not, by itself, invert the stereogenic carbon because its C–O bond remains intact. A later SN2 displacement at carbon can invert it. This two-step reasoning separates the site of reagent attack from the stereocentre whose geometry is being tracked.
The Lucas-test idea uses conversion of suitable alcohols to less water-soluble alkyl chlorides under strongly acidic activating conditions. Tertiary substrates often react rapidly, secondary more slowly and ordinary primary ones slowly under the usual comparison. Benzylic/allylic reactivity and solubility can be exceptions, so this is a conditional qualitative test rather than an infallible structural proof. No handling procedure is needed to understand that reasoning.
07 / UNDERSTAND & APPLY
7. Elimination competes with substitution and can rearrange a skeleton
Acid-catalysed dehydration removes the elements of water to form an alkene. For secondary and tertiary alcohols, a common pathway protonates OH, loses water to form a carbocation and removes a beta hydrogen. The acid is regenerated. A primary substrate need not form a free primary carbocation; its elimination can have a more concerted character.
The OH-bearing carbon is the alpha carbon for elimination, and adjacent carbons are beta carbons. A beta hydrogen and a bond to the leaving group are required for the usual alkene-forming process. List distinct beta positions before deciding which constitutional alkenes are possible.
More substituted alkenes are often favoured under ordinary acid dehydration conditions, and E alkenes can be favoured over Z when their steric environments differ. These are trends, not substitutes for mechanism. Conjugation, ring constraints, rearrangement and equilibration can alter the product distribution.
A carbocation may undergo a hydride or alkyl shift if a more favourable cation is accessible. The migrating bond brings its electron pair to the cationic centre; the positive charge appears on the carbon that lost that bond. Carbon count is conserved even when the skeleton is rearranged.
Phosphoryl chloride with a suitable base is a standard conceptual route for alcohol dehydration through activation followed by E2-type elimination, avoiding a free-carbocation rearrangement pathway in suitable substrates. Such a route still requires compatible geometry and beta hydrogen. Distinguish the mechanism from acid-catalysed E1 reasoning.
Intermolecular dehydration of suitable primary alcohols can form ethers under selected conditions, while intramolecular elimination gives alkenes. Temperature, substrate and acid conditions influence competition. A secondary or tertiary alcohol is not generally a clean choice for a simple symmetrical-ether preparation because elimination can dominate.
ORIGINAL WORKED EXAMPLE
Enumerate dehydration products
For pentan-2-ol, list the constitutional alkene products available by eliminating a beta H without rearrangement, and identify the usual substitution trend.
The OH-bearing alpha carbon is C2. Beta carbons are C1 and C3.
Removing H from C1 gives pent-1-ene; removing H from C3 gives pent-2-ene.
Pent-2-ene can have E and Z forms.
Under ordinary alkene-stability-controlled conditions the internal alkene is often favoured, with E often favoured over Z, but exact ratios require conditions.
Answer Pent-1-ene and E/Z-pent-2-ene; the more substituted pent-2-ene is commonly favoured.
08 / UNDERSTAND & APPLY
8. Reactions at oxygen need not replace the carbon–oxygen bond
Alcohols react with sufficiently strong bases to form alkoxides. Reaction with an active metal can be summarized as 2ROH+2Na→2RONa+H₂. Each OH supplies one proton, so one mole of a monohydric alcohol can produce one-half mole of hydrogen in the ideal complete stoichiometric conversion. This is an atom-balance relationship, not a recommendation for an experiment.
Alkoxides are useful nucleophiles in Williamson ether synthesis, especially with methyl or primary alkyl electrophiles compatible with SN2. A hindered tertiary alkyl halide instead tends to undergo elimination with a strong alkoxide base. Choose which fragment provides the alkoxide and which provides the leaving-group-bearing carbon to minimize that problem.
An alcohol reacts with a carboxylic acid in Fischer esterification to form an ester and water under acid catalysis. The alcohol oxygen normally becomes the single-bond ester oxygen in the ordinary acyl-substitution mechanism. Carboxylic-acid oxygen exchange under acidic conditions can complicate isotope experiments, so an atom-label prediction needs explicit reaction assumptions.
Acyl chlorides and anhydrides can also form esters from alcohols through nucleophilic acyl substitution. This is different from converting the alcohol carbon into a halide: the alcohol C–O bond is retained while oxygen bonds to an acyl carbon. Acidic or basic by-products and other functional groups affect practical compatibility.
An OH group can be protected as a suitable ether, such as a silyl ether, when its acidity or nucleophilicity would interfere with a later step. Protection, the intended transformation and deprotection must all be compatible with the rest of the molecule. A protecting group is not a universal shield against every reagent, and unnecessary protection adds steps and losses.
ORIGINAL WORKED EXAMPLE
Hydrogen stoichiometry from OH count
A pure sample contains 0.0600 mol of a diol with two reactive OH groups per molecule. In an ideal complete conversion to its dialkoxide by excess active metal, how much H₂ is formed?
There are 2×0.0600=0.120 mol OH groups.
Each two OH protons produce one H₂ molecule in the balanced conversion.
Thus n(H₂)=0.120/2=0.0600 mol.
Answer 0.0600 mol H₂, assuming both OH groups react completely.
09 / UNDERSTAND & APPLY
9. Diols create additional reaction possibilities
A vicinal diol has OH groups on adjacent carbons. Oxidative cleavage with a suitable periodate-type reagent can break the C–C bond between those centres through a cyclic intermediate when the required geometry is accessible. Each original OH-bearing carbon becomes a carbonyl centre, so its remaining substituents determine whether an aldehyde or ketone is formed.
For HOCH₂–CH(OH)–CH₂CH₃, cleavage of the bond between the two OH-bearing carbons gives methanal from CH₂OH and propanal from the other carbon. Count all carbon atoms: one plus three equals the original four. Do not cut every C–C bond in the molecule or insert an extra carbon at the new aldehyde group.
Acid-catalysed pinacol rearrangement is a different process: a suitable diol loses water, then a neighbouring group migrates with accompanying carbonyl formation. It does not simply split the molecule into two carbonyl fragments. Product prediction depends on which group leaves, cation stabilization, migratory possibilities and geometry; a single universal migration order is not adequate for every substrate.
To design a multistep sequence, mark whether each step changes carbon count, oxidation level or connectivity at oxygen. For example, alkene hydration followed by secondary-alcohol oxidation gives a ketone, while Grignard addition followed by workup increases carbon count and gives a different alcohol class. Writing the intermediate structure avoids treating reagent lists as unconnected codes.
A route can fail because a reagent reacts with an earlier functional group. Oxidation before carbon–carbon addition may be useful; unprotected OH during Grignard chemistry is a problem. Also check stereochemistry after a planar carbonyl intermediate or free carbocation, because a previous stereocentre may be erased or rearranged.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Product-connectivity diagram, not a complete periodate mechanism. Suitable geometry and conditions for the stated oxidative cleavage are assumed. This cleavage differs from a pinacol rearrangement, which migrates a group rather than simply producing these two fragments.
ORIGINAL WORKED EXAMPLE
Track a vicinal-diol cleavage
Predict products from ideal periodate cleavage of CH₃CH(OH)CH(OH)CH₂CH₃, assuming the cyclic intermediate can form.
Break only the bond between the two OH-bearing carbons, C2 and C3.
C2 retains CH₃ and H, so it becomes CH₃CHO.
C3 retains CH₂CH₃ and H, so it becomes CH₃CH₂CHO.
The products contain 2+3=5 carbons, equal to the starting diol.
Answer Ethanal and propanal.
10 / UNDERSTAND & APPLY
10. Solved practice and a bond-tracking audit
For each proposed answer, check the OH-bearing carbon’s neighbours, every oxygen-containing fragment and the total carbon count. State whether the pathway uses an activated leaving group, a planar carbonyl or a carbocation. These decisions determine more than a memorized primary/secondary/tertiary label alone.
All exercises here are original teaching cases. Reagent descriptions identify conceptual transformations and selectivity, not laboratory handling procedures.
ORIGINAL PRACTICE · 1
Classify 2,2-dimethylpropan-1-ol as primary, secondary or tertiary.
Reveal answer & reasoning
Primary.
Its OH-bearing CH₂ is bonded to only one other carbon. The neighbouring carbon is highly branched, but that does not change the alcohol class.
ORIGINAL PRACTICE · 2
What alcohol is formed by ordinary NaBH₄ reduction of heptan-3-one followed by workup?
Reveal answer & reasoning
Heptan-3-ol.
Hydride adds at the carbonyl carbon without changing the seven-carbon skeleton. In an achiral environment a new chiral C3 gives a racemic product absent other influences.
ORIGINAL PRACTICE · 3
Can ordinary reduction of a ketone directly make a tertiary alcohol?
Reveal answer & reasoning
No.
Adding H to the carbonyl carbon produces an OH-bearing carbon with a hydrogen and two carbon substituents, which is secondary.
ORIGINAL PRACTICE · 4
What results from ethylmagnesium halide adding to formaldehyde, then workup?
Reveal answer & reasoning
Propan-1-ol.
The reagent’s two-carbon ethyl group gains the one carbon of formaldehyde, giving CH₃CH₂CH₂OH.
ORIGINAL PRACTICE · 5
What results from methylmagnesium halide adding to butanal, then workup?
Reveal answer & reasoning
Pentan-2-ol.
The former aldehyde carbon retains H and propyl while gaining methyl and OH. The total carbon count is 4+1=5.
ORIGINAL PRACTICE · 6
Why does an unprotected OH group interfere with a Grignard reagent?
Reveal answer & reasoning
It protonates the strongly basic carbon nucleophile.
RMgX can be converted to RH by acid–base reaction, consuming reagent before or alongside the desired carbonyl addition.
ORIGINAL PRACTICE · 7
What is the ordinary oxidation product of cyclopentanol?
Reveal answer & reasoning
Cyclopentanone.
The OH-bearing ring carbon is secondary and has a hydrogen that can be removed in carbonyl formation.
ORIGINAL PRACTICE · 8
Why does 2-methylbutan-2-ol resist ordinary alcohol-to-carbonyl oxidation?
Reveal answer & reasoning
Its OH-bearing carbon has no hydrogen.
Making a carbonyl there while retaining all three C–C bonds would exceed carbon valence; harsher cleavage chemistry is a different process.
ORIGINAL PRACTICE · 9
Does tosylating an enantiopure secondary alcohol by reaction at oxygen itself invert the alcohol carbon?
Reveal answer & reasoning
No.
The C–O bond remains intact during oxygen sulfonylation. A later SN2 substitution at carbon can produce inversion.
ORIGINAL PRACTICE · 10
Which Williamson disconnection is better for methyl tert-butyl ether: tert-butoxide with a methyl electrophile, or methoxide with a tertiary electrophile?
Reveal answer & reasoning
tert-Butoxide with a suitable methyl electrophile.
SN2 attack at methyl is feasible; a tertiary electrophile is hindered and promotes elimination with strong alkoxide.
ORIGINAL PRACTICE · 11
How much H₂ is formed ideally from 0.100 mol of a monohydric alcohol converted completely to alkoxide by active metal?
Reveal answer & reasoning
0.0500 mol.
The balanced equation uses two ROH per H₂, so divide the alcohol amount by two.
ORIGINAL PRACTICE · 12
Does periodate cleavage of a vicinal diol generally have the same connectivity outcome as pinacol rearrangement?
Reveal answer & reasoning
No.
Cleavage splits the bond between two OH-bearing carbons into separate carbonyl fragments; rearrangement migrates a group and forms a carbonyl without that simple two-fragment pattern.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.