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(A) The reaction involves the hydroboration-oxidation of methylenecyclohexane. $(i)$ The first step is the addition of borane $(BH_3)_2$ across the do

Vedclass · Accepted Answer

Cyclohexylmethanol

Vedclass · Answer

(A) The reaction involves the hydroboration-oxidation of methylenecyclohexane. $(i)$ The first step is the addition of borane $(BH_3)_2$ across the double bond,which follows anti-Markovnikov addition. (ii) The second step is the oxidation with $H_2O_2$ in the presence of $OH^-$,which replaces the boron atom with a hydroxyl group $(-OH)$. This process results in the formation of a primary alcohol. Thus,methylenecyclohexane reacts to form cyclohexylmethanol.

The product $(P)$ in the given reaction is:

Similar Questions

In the reaction $CH_3-CH=CH_2 + H_2O + [O]$ $\xrightarrow[Acid]{KMnO_4} CH_3-CH(OH)-CH_2(OH)$ $\xrightarrow{[O]} X + HCOOH$,$X$ is:

Predict the major product for the following reaction:
$CH_2=CH-CH_2-I \xrightarrow{HI \text{ (Excess)}} ?$

$HO - CH_2 - CH_2 - CH_2 - CH_2 - CH = CH_2 \xrightarrow{Br_2} \text{Product ?}$

$1, 2$-dibromocyclohexane on dehydrohalogenation gives

$CH_2 = CH_2 \xrightarrow[KOH / H_2O]{KMnO_4} X$. The product $X$ in the above reaction is:

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