Determine the correct configuration (E or Z) | vedclass

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(B) To determine the $E/Z$ configuration,we use the Cahn-Ingold-Prelog $(CIP)$ priority rules for the groups attached to each carbon of the double bon

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$Z$

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(B) To determine the $E/Z$ configuration,we use the Cahn-Ingold-Prelog $(CIP)$ priority rules for the groups attached to each carbon of the double bond.$1$. For the left carbon $(C_1)$: The groups are $-CH=NH$ and $-NH-CH_3$. Comparing atomic numbers of the first atoms,$N$ (atomic number $7$) has higher priority than $C$ (atomic number $6$). Thus,$-NH-CH_3$ has higher priority.$2$. For the right carbon $(C_2)$: The groups are $-CH_2-NH_2$ and $-C \equiv N$. Comparing the first atoms,both are $C$. Looking at the next atoms: $-CH_2-NH_2$ is bonded to $(H, H, N)$ while $-C \equiv N$ is bonded to $(N, N, N)$ (treating triple bonds as three bonds to $N$). Since $N$ has a higher atomic number than $H$,the $-C \equiv N$ group has higher priority.$3$. In the given structure,the higher priority groups ($-NH-CH_3$ on the left and $-C \equiv N$ on the right) are on the same side of the double bond.$4$. When the higher priority groups are on the same side,the configuration is $Z$ (from the German 'zusammen').

Determine the correct configuration ($E$ or $Z$) for the following compound:
$HN=CH-C(CH_3-NH)=C(CH_2-NH_2)-C \equiv N$

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Determine the correct configuration ($E$ or $Z$) for the following compound:$HN=CH-C(CH_3-NH)=C(CH_2-NH_2)-C \equiv N$