In the reaction shown below,the o/p ratio wil | vedclass

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(A) The $o/p$ ratio refers to the ratio of the ortho-substituted product to the para-substituted product. As the size of the substituent $R$ increases

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$R=-CH_3$

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(A) The $o/p$ ratio refers to the ratio of the ortho-substituted product to the para-substituted product. As the size of the substituent $R$ increases,the steric hindrance at the ortho position increases,which makes the formation of the ortho-product more difficult compared to the para-product.Therefore,the $o/p$ ratio is highest when the substituent $R$ is the smallest.Comparing the sizes of the given alkyl groups:$R=-CH_3$ (smallest)$R=-CH_2-CH_3$$R=-CHMe_2$$R=-CMe_3$ (largest)Since $-CH_3$ is the smallest group,it offers the least steric hindrance to the incoming electrophile at the ortho position,resulting in the highest $o/p$ ratio.$R$$\% o-$$\% p-$$CH_3$$58$$37$$CH_2Me$$45$$49$$CHMe_2$$30$$62$$CMe_3$$16$$73$

In the reaction shown below,the $o/p$ ratio will be highest when:
$C_6H_5R + HNO_3 \xrightarrow{H_2SO_4} o-\text{nitro-substituted product} + p-\text{nitro-substituted product}$

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In the reaction shown below,the $o/p$ ratio will be highest when:$C_6H_5R + HNO_3 \xrightarrow{H_2SO_4} o-\text{nitro-substituted product} + p-\text{nitro-substituted product}$