JEE Main 20204 SeptemberEvening Shift
- Exam
- JEE Main
- Year
- 2020
- Date
- 4 September
- Shift
- Evening Shift
- Subject
- Chemistry
- Chapter
- Aldehydes and Ketones
JEE Main 2020 · 4 September · Evening Shift · Chemistry · Aldehydes and Ketones
The major product [ C ] of the following reaction sequence will be :
The major product of the following reaction sequence will be :
Options
- A
![Consider the following low-spin complexes K_3[Co(NO_2)_6], K_4[Fe(CN)_6] , K_3[Fe(CN)_6], Cu u_2[Fe(CN)_6] and](/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fjee_main%252Fed465e2b-7cb4-4d41-8e86-07937c8f2c48-image.png%3Falt%3Dmedia&clean=1&v=qxfig138)
- B
![Consider the following low-spin complexes K_3[Co(NO_2)_6], K_4[Fe(CN)_6] , K_3[Fe(CN)_6], Cu u_2[Fe(CN)_6] and](/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fjee_main%252F337991b7-57e9-4f11-b8e7-d1d43a27137c-image.png%3Falt%3Dmedia&clean=1&v=qxfig138)
- C
![Consider the following low-spin complexes K_3[Co(NO_2)_6], K_4[Fe(CN)_6] , K_3[Fe(CN)_6], Cu u_2[Fe(CN)_6] and](/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fjee_main%252F83557426-e94d-4adc-89cc-b03e9b15afbb-image.png%3Falt%3Dmedia&clean=1&v=qxfig138)
- D
![Consider the following low-spin complexes K_3[Co(NO_2)_6], K_4[Fe(CN)_6] , K_3[Fe(CN)_6], Cu u_2[Fe(CN)_6] and](/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fjee_main%252Fee4f95f9-cd3b-4954-a095-7dbcacebe08e-image.png%3Falt%3Dmedia&clean=1&v=qxfig138)
Correct answer
A. ![Consider the following low-spin complexes K_3[Co(NO_2)_6], K_4[Fe(CN)_6] , K_3[Fe(CN)_6], Cu u_2[Fe(CN)_6] and](/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fjee_main%252Fed465e2b-7cb4-4d41-8e86-07937c8f2c48-image.png%3Falt%3Dmedia&clean=1&v=qxfig138)
Step-by-step solution
Markovnikov's addition states that, in hydrohalogenation of an unsymmetrical alkene, the hydrogen atom in the hydrogen halide forms a bond with the doubly bonded carbon atom in the alkene, bearing the greater number of hydrogen atoms.