The absolute configurations of the two chiral | vedclass

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

Question

(A) To determine the absolute configuration $(R/S)$ at each chiral center,we assign priorities to the groups attached to each carbon using the Cahn-In

Vedclass · Accepted Answer

$2(R), 3(S)$

Vedclass · Answer

(A) To determine the absolute configuration $(R/S)$ at each chiral center,we assign priorities to the groups attached to each carbon using the Cahn-Ingold-Prelog $(CIP)$ rules.For carbon-$2$:Priorities: $1: -Br$,$2: -COOH$,$3: -CH(OH)CN$,$4: -H$.The sequence $1$ $\rightarrow 2$ $\rightarrow 3$ is counter-clockwise. Since the lowest priority group $(-H)$ is on a horizontal bond,the configuration is reversed from $S$ to $R$. Thus,carbon-$2$ is $2(R)$.For carbon-$3$:Priorities: $1: -OH$,$2: -CH(Br)COOH$,$3: -CN$,$4: -H$.The sequence $1$ $\rightarrow 2$ $\rightarrow 3$ is clockwise. Since the lowest priority group $(-H)$ is on a horizontal bond,the configuration is reversed from $R$ to $S$. Thus,carbon-$3$ is $3(S)$.Therefore,the configuration is $2(R), 3(S)$.

The absolute configurations of the two chiral centers in the following molecule are:

Similar Questions

Which of the following possesses a dome-shaped thalamus?

The maximum area of a right-angled triangle with hypotenuse $h$ is

$A$ plane passing through $(-1, 2, 3)$ and whose normal makes equal angles with the coordinate axes is

$A$ straight conductor of length $4 ~m$ moves at a speed of $10 ~m/s$. When the conductor makes an angle of $30^{\circ}$ with the direction of the magnetic field of induction $0.1 ~Wb/m^2$,the induced emf is: (in $~V$)

The mirror image of the point $P(1, 2, 3)$ in a plane is $P'\left(-\frac{7}{3}, -\frac{4}{3}, -\frac{1}{3}\right)$. Which of the following points lies on this plane?

Vedclass Test Series

Exam Paper Generator

Online Exam Module