Question

How many isomeric alcohols (including stereoisomers) with molecular formula C$_{5}$H$_{12}$O on dehydration and reduction with hydrogen on palladium catalyst form same hydrocarbon (D) C$_{5}$H$_{12}$. Hydrocarbon D on monochlorination form four constitutional isomers of monochloro product.

Answer 1

8

Answer 2

We first note that D must be the C₅H₁₂ which—by monochlorination—yields exactly 4 constitutional chlorides. Checking the three pentanes:

• n‑Pentane gives only 3 (1‑, 2‑, and 3‑chloropentane).

• Neopentane (2,2‑dimethylpropane) gives only 1.

• 2‑Methylbutane (isopentane) gives 4 different types because when you analyze the different types of H (using its structure:

  CH₃–CH(CH₃)–CH₂–CH₃),

  you find that the hydrogens fall into 5 types:

  • on the left terminal CH₃ group (call it A),
  • on the CH (B) which has one H,
  • on the CH₂ (C),
  • on the right terminal CH₃ (D),
  • and on the methyl branch attached to B (E).

However, on close comparison, the H’s on carbon A and on the branch (E) turn out to be equivalent in connectivity to the parent hydrocarbon (both are –CH₃ groups attached to the same carbon B), so they give the same hydrocarbon when “OH” is removed. Thus, effectively there are 4 distinct carbon sites (A/E counting as one unique type, B, C, and D) on 2‑methylbutane corresponding to the 4 types of monochloro products.

Now, if an alcohol (C₅H₁₂O) gives D (2‑methylbutane) on dehydration (or reduction by H₂/Pd), then it must be derived from 2‑methylbutane by “inserting” an –OH group at one of the H–sites. When we “replace” one H by –OH in 2‑methylbutane, there are 5 sets of H’s (as seen above: on A, B, C, D, and E). But note that although hydrogen atoms on A and on the branch (E) are equivalent in D, an –OH group attached at the main‐chain terminal versus attached as a branch leads to different constitutional connectivity in the alcohol.

Thus we list the possibilities:

1. –OH at carbon A (the left terminal CH₃): gives an alcohol that, when renumbered properly, is 2‑methylbutan‑1‑ol.
2. –OH at carbon D (the right terminal CH₃): gives a different alcohol (named by numbering from the other end).
3. –OH at carbon B (the CH that normally has one H): substitution here produces a new chiral centre hence a pair of enantiomers.
4. –OH at carbon C (the CH₂): substitution gives a chiral centre, so 2 stereoisomers (a pair of enantiomers).
5. –OH at carbon E (the methyl branch on C₂): substitution here also gives a chiral centre (when considered in the overall connectivity) and produces 2 enantiomers.

Counting constitutional isomers together with their stereoisomers:

• From A: 1 isomer
• From D: 1 isomer
• From B: 2 stereoisomers
• From C: 2 stereoisomers
• From E: 2 stereoisomers

Total = 1 + 1 + 2 + 2 + 2 = 8.

Thus, 8 isomeric alcohols (including stereoisomers) with formula C₅H₁₂O will, on dehydration or catalytic hydrogenolysis, yield the same hydrocarbon D (2‑methylbutane).

Explanation (minimal):

1. Identify D as 2‑methylbutane because its monochlorination affords 4 constitutional isomers.
2. Recognize that alcohols with formula C₅H₁₂O leading to D come from inserting –OH into 2‑methylbutane.
3. There are 5 different sets of H (one on each of A, B, C, D, and E) but the sites on A and on the branch (E) give different alcohols.
4. Replacements at the secondary centers (B, C, and E) produce chiral centres, each giving 2 stereoisomers.
5. Total count = 1 (A) + 1 (D) + 2 (B) + 2 (C) + 2 (E) = 8.