Question

${{\left( C{{H}_{3}} \right)}_{3}}COH\xrightarrow{573K,Cu}A\xrightarrow{{{H}_{2}}{{O}_{2}}/OH/{{B}_{2}}{{H}_{6}}/THF}B$ . B is:
[A] $C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OH$
[B] $C{{H}_{3}}-CH\left( C{{H}_{3}} \right)-C{{H}_{2}}-OH$
[C] $C{{H}_{3}}-CHOH-C{{H}_{2}}-C{{H}_{3}}$
[D] $C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{3}}$

Answer 1

HINT: To solve this proceed step-wise. Remember that tertiary alcohols undergo dehydration in presence of a metal like copper and high temperature. From this you can find A. Also remember that in presence of hydrogen peroxide, the addition is Anti- Markovnikov's.

COMPLETE ANSWER: To solve this and to find B, let us proceed step wise and find A first and then we can find B. Firstly, we can see that we have a tertiary alcohol as the reactant here. A tertiary alcohol is a compound where there are no hydrogen substituents on the carbon atom. Its general formula is $C{{R}_{2}}OH$. Here, we have ${{\left( C{{H}_{3}} \right)}_{3}}COH$. With Copper metal and at high temperature as 573K, tertiary alcohols undergo dehydration. Here, the product obtained will be 2 – methylprop – 1 – ene.
We can write the reaction as-
\[{{(C{{H}_{3}})}_{3}}COH\xrightarrow{573K,Cu}{{(C{{H}_{3}})}_{2}}C=C{{H}_{2}}\]
Now, to this, diborane and hydrogen peroxide are added in THF as the solvent. Here, the reaction will be an additional reaction.
We know that in presence of hydrogen peroxide, addition takes place according to anti – Markonikov’s rule due to the peroxide effect. As a result, the hydroxyl group will be added to least substituted carbon atom i.e. the carbon atom across the double bond with the most number of hydrogen atoms. So, we can write the reaction here as-
\[{{(C{{H}_{3}})}_{2}}C=C{{H}_{2}}\xrightarrow{{{H}_{2}}{{O}_{2}}/OH/{{B}_{2}}{{H}_{6}}/THF}{{(C{{H}_{3}})}_{2}}CH-C{{H}_{2}}-OH\]

Therefore, we can understand that the correct answer is option [B] $C{{H}_{3}}-CH\left( C{{H}_{3}} \right)-C{{H}_{2}}-OH$

NOTE: Here, if we had a secondary or a primary alcohol instead of a tertiary alcohol, then the reaction would be dehydrogenation and not dehydration in addition to Cu at high temperature as the above reaction. A primary alcohol would lead to the formation of an aldehyde and a secondary alcohol would give us a ketone and tertiary alcohol as we have already discussed above, and gives an alkene.