Answer
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Hint: In such questions where aldehyde, nitro and methoxy groups are attached in a compound then the aldehyde is given the highest preference. Therefore, the numbering will start from the aldehyde group.
Complete step by step answer:
In this question we have three groups attached to the benzene ring. These are $CHO$, $OC{H_3}$ and $N{O_2}$. The $CHO$ is the benzaldehyde group , $OC{H_3}$ is called the methoxy group and $N{O_2}$ is called the nitro group. For naming the compounds the decreasing order of preference according to $IUPAC$ can be given as:
Carboxylic acid $ > $ sulfonic acid $ > $ ester $ > $ halide $ > $ amide $ > $ ketone $ > $ alcohol $ > $ aldehyde $ > $ ketone $ > $ alcohol $ > $ thiol $ > $ amine $ > $ alkene $ > $ alkyne $ > $ alkane $ > $ ether(alkoxy) $>$ nitro.
Here, it can be seen that aldehyde has the most preference, then the methoxy group that is an ether and finally the nitro group. Therefore, we will start numbering from carbon which consist of the aldehyde group and then the methoxy group will get number$ - 2$ and the nitro group will get number$ - 4$. This numbering can be shown as:
The prefix for the given compound will be methoxy and nitro, and the suffix will be aldehyde. Here, the suffix will be named in the alphabetical order.
Therefore, the name of the given compound can be written as:
$\text{2 - methoxy - 4 - nitro benzaldehyde}$
Hence, option A is the correct answer.
Additional information:
Some functional groups like halides, nitro, alkoxide have no suffixes for them . Therefore, they are used as prefixes while naming the compounds. Also, the functional group which has the highest priority in the given compound will always be used as a suffix while naming the compounds.
Note: Always remember that while doing $IUPAC$ naming the compound which has the most preference will be chosen as a suffix in the naming and the other compounds will be arranged alphabetically in naming.
Complete step by step answer:
In this question we have three groups attached to the benzene ring. These are $CHO$, $OC{H_3}$ and $N{O_2}$. The $CHO$ is the benzaldehyde group , $OC{H_3}$ is called the methoxy group and $N{O_2}$ is called the nitro group. For naming the compounds the decreasing order of preference according to $IUPAC$ can be given as:
Carboxylic acid $ > $ sulfonic acid $ > $ ester $ > $ halide $ > $ amide $ > $ ketone $ > $ alcohol $ > $ aldehyde $ > $ ketone $ > $ alcohol $ > $ thiol $ > $ amine $ > $ alkene $ > $ alkyne $ > $ alkane $ > $ ether(alkoxy) $>$ nitro.
Here, it can be seen that aldehyde has the most preference, then the methoxy group that is an ether and finally the nitro group. Therefore, we will start numbering from carbon which consist of the aldehyde group and then the methoxy group will get number$ - 2$ and the nitro group will get number$ - 4$. This numbering can be shown as:
The prefix for the given compound will be methoxy and nitro, and the suffix will be aldehyde. Here, the suffix will be named in the alphabetical order.
Therefore, the name of the given compound can be written as:
$\text{2 - methoxy - 4 - nitro benzaldehyde}$
Hence, option A is the correct answer.
Additional information:
Some functional groups like halides, nitro, alkoxide have no suffixes for them . Therefore, they are used as prefixes while naming the compounds. Also, the functional group which has the highest priority in the given compound will always be used as a suffix while naming the compounds.
Note: Always remember that while doing $IUPAC$ naming the compound which has the most preference will be chosen as a suffix in the naming and the other compounds will be arranged alphabetically in naming.
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