c) 2 moles of R + 2 moles of S = 0 ( each will cancel out) So only remaining 6 m

Question

c)

2 moles of R + 2 moles of S = 0 ( each will cancel out)

So only remaining 6 moles of S isomer will rotate polarised light among 10 moles of total compounds.

e.e % = ( 6/10) * 100 => 60 % (option d)

d) meso compound will not rotate plane polarised light.

Answer option b) the second molecule which has two Br that is one pointed above the plane(darken line) the second pointed below the plane ( dashed line)

e) option a) secondary amine ( cyclopentamine)

H bonded with N of one molecule can interact with lone pair of N atom in other amine can interact.

Explanation / Answer

(c) What is the enantiomeric excess (e.e.) of a mixture of 2 moles of R-2-butanol and 8 moles S- 2-butanol? 30% 40% 50% GY% 70% (d) Which of the following compounds will not rotate the plane of polarized light? Br Br Br Br (e) Which of the following molecules can act as both a hydrogen bond donor and acceptor? , (D How does the rate of hydrolysis of ter-butyl bromide change when concentration of tert-buty bromide is doubled?

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