The product of the reaction is:

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given reactant is a diol containing a primary alcohol $(-CH_2OH)$ and a tertiary alcohol $(-C(OH)(CH_3)_2)$.Pyridinium chlorochromate $(PCC)$

Vedclass · Accepted Answer

$4-$($2$-hydroxypropan$-2-$yl)cyclohexanecarbaldehyde

Vedclass · Answer

(B) The given reactant is a diol containing a primary alcohol $(-CH_2OH)$ and a tertiary alcohol $(-C(OH)(CH_3)_2)$.Pyridinium chlorochromate $(PCC)$ is a selective oxidizing agent that oxidizes primary alcohols to aldehydes and secondary alcohols to ketones.It does not oxidize tertiary alcohols.Therefore,the primary alcohol group is oxidized to an aldehyde group $(-CHO)$,while the tertiary alcohol group remains unchanged.The product is $4-(2-hydroxypropan-2-yl)cyclohexanecarbaldehyde$.

Similar Questions

What is the final product (major) $A$ in the given reaction?

When wine is exposed to air,it becomes sour. The reason for this is -

Which of the following alcohols gives white turbidity almost immediately with the Lucas reagent at room temperature?
$(i)$ $n-$Butanol
$(ii)$ $tert-$Butanol
$(iii)$ Benzyl alcohol
$(iv)$ Allyl alcohol

Iodoform can be obtained on warming $NaOH$ and iodine with:

Reactivity of given alcohols for acidic dehydration can be explained by $Ph-CH_2-CH(OH)-CH_3 > CH_3-CH_2-CH(OH)-CH_3$:

Vedclass Test Series

Exam Paper Generator

Online Exam Module