Sum of number of 1^o alcoholic groups in prod | vedclass

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(D) $1$. In the first reaction,$NaBH_4$ is a selective reducing agent that reduces aldehydes and ketones but does not reduce esters. The starting mate

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$5$

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(D) $1$. In the first reaction,$NaBH_4$ is a selective reducing agent that reduces aldehydes and ketones but does not reduce esters. The starting material has an aldehyde group,a ketone group,and an ester group. The aldehyde and ketone are reduced to $1^o$ and $2^o$ alcohols respectively,while the ester remains unchanged. However,the provided solution image indicates a product $(P)$ where all groups are reduced to alcohols,which implies a stronger reducing agent like $LiAlH_4$ was intended or the product structure is as shown. Based on the provided solution image,$(P)$ contains two $1^o$ alcoholic groups.$2$. In the second reaction,$LiAlH_4$ is a strong reducing agent that reduces both the lactone (cyclic ester) and the aldehyde group. The reduction of the lactone opens the ring to form a diol,and the aldehyde is reduced to a primary alcohol. Based on the provided solution image,$(Q)$ contains three $1^o$ alcoholic groups.$3$. The sum of $1^o$ alcoholic groups in $(P)$ and $(Q)$ is $2 + 3 = 5$.

Column-$I$ (Reaction)	Column-$II$ (Bond broken)
$(A)$ $RCH_2OH \xrightarrow{CrO_3} RCHO$	$(i)$ $C-O$ bond breaks.
$(B)$ Phenol $\xrightarrow{HNO_3}$ $p$-Nitrophenol + $o$-Nitrophenol	(ii) $O-H$ bond breaks.
$(C)$ Phenol + $NaOH \rightarrow$ Sodium phenoxide + $H_2O$	(iii) $O-H$ and $C-H$ bonds break.
$(D)$ $CH_3CH_2CH_2OH \xrightarrow{H_2SO_4, \Delta} CH_3CH=CH_2$	(iv) $C-H$ bond breaks.

Sum of number of $1^o$ alcoholic groups in product $(P)$ and $(Q)$ is

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