Match the reactions (List-I) with reagents (List-II) used to get the product.
| LIST-I (Reactions) | LIST-II (Reagents) | ||
| A. |  | I. | H+/ H2O |
| B. |  | II. | BH3/ H2O2, OH− |
| C. |  | III. | Lithium aluminum hydride/ H2O |
| D. |  | IV. | Sodium borohydride |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-III, B-I, C-IV, D-II
Alcohols are made in two main ways. One is reducing a C=O group, where \( \ce{NaBH4} \) is the milder reagent and \( \ce{LiAlH4} \) the stronger one. The other is adding water across a C=C bond, where acid-catalysed hydration follows Markovnikov’s rule and hydroboration–oxidation gives the anti-Markovnikov product.
A. Aldehyde → primary alcohol — This matches III. Lithium aluminum hydride/ H2O. \( \ce{LiAlH4} \) reduces –CHO to –\( \ce{CH2OH} \); A–III follows by elimination, because the products of B, C and D fix their reagents.
\( \displaystyle \ce{R-CHO} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{LiAlH4}} \ce{R-CH2OH} \)
B. 1-Methylcyclopentene → 1-methylcyclopentanol — This matches I. H+/ H2O. In acid-catalysed hydration, \( \ce{H^{+}} \) adds to the less substituted carbon and forms the more stable tertiary carbocation, so the –OH ends up on the carbon carrying the methyl group (Markovnikov addition).
\( \displaystyle \text{1-methylcyclopentene} \xrightarrow{\ce{H^+}/\ce{H2O}} \text{1-methylcyclopentanol} \)
C. Cyclopent-2-en-1-one → cyclopent-2-en-1-ol — This matches IV. Sodium borohydride. The C=O is conjugated with the C=C. The mild reagent \( \ce{NaBH4} \) reduces only the C=O and leaves the double bond intact.
\( \displaystyle \text{cyclopent-2-en-1-one} \xrightarrow{\ce{NaBH4}} \text{cyclopent-2-en-1-ol} \)
D. 1-Methylcyclopentene → 2-methylcyclopentanol — This matches II. BH3/ H2O2, OH−. Boron adds to the less substituted carbon and is then replaced by –OH, so the –OH ends up on the carbon next to the methyl group (anti-Markovnikov addition).
\( \displaystyle \text{1-methylcyclopentene} \xrightarrow[\text{(ii) }\ce{H2O2},\ \ce{OH^-}]{\text{(i) }\ce{BH3}} \text{2-methylcyclopentanol} \)
Shortcut: A–III appears only in Options (1) and (4), and C–IV then leaves only Option (4).
Note: It seems that there is a mistake in the original NTA paper. In reaction A, the product is drawn with a saturated ring (cyclopentylmethanol); it should keep the ring double bond (cyclopent-3-en-1-ylmethanol), because \( \ce{LiAlH4} \) reduces the C=O group but does not reduce an isolated C=C bond.