Q6; More triethylamine can dissolve be dissolved if more WATER is added. Hence o

Question

Q6;

More triethylamine can dissolve be dissolved if more WATER is added. Hence option (b) is correct answer. Triethyl amine is a base, if we add acid like H2SO4 as is given in option (c) a neutralisation reaction will take place , as a result more base will be dissolved in water (dissolution equilibrium will be shifted to right direction). So option (c) is correct answer. Further adding more water will also make triethyl amine more soluble in water. So if you have to give more than one option then option (b) and option (c) is correct.

Q7.

The pure -R has negative specific rotation. The mixture has a positive specific rotation, The mixture must be ritcher in S-enantionmer.

Enantiomeric excess (of S enatiomer) = (Observd rotation of the mixture / speific rotation) X 100

= (+0.500 / + 40.0) X 100 = 1.25%

Hence isomer S is in 1.25% excess and remaining (100 - 1.25) = 98.75% is R/S mixture.

Amount of R = 98.75 / 2 = 49.375%

Amount of S = 100 - 49.375 = 50.625%

But there is no option in tune with this result. So may be the question is wrong.

Q8. Option (a) is the correct answer. As the compound has an internal plan of symmetry so the rotation of one half is cancelled by the other half.

Explanation / Answer

6. What would make triethylamine (CH3CH2)3N more soluble in water? a. adding NaOH b. adding more water c. adding 2SO4 d. adding hexane e. adding NaHCO 7. The optical rotation of the pure R-enantiomer ofa an optical rotation of [+0.500° for a solution of 0.0250 g/mL in a ldm tube. What is the composition of the sample? a. 75 % R and 25 %S b, 75 % S and 25 % R c. 1.25 % and 98.75 % S d, 1.25 %S and 98.75 % R e. 7.5 % R and 92.5 % 8. Which of the following compounds is a meso compound? a. (2R,3R)-dibromobutane b. (2R,3S)-dibromobutane c. (2R,3S)-3-bromo-2-iodobutane d· (2R,3R)-3-bromo-2-iodobutane e. (2S,3R)-3-bromo-2-iodobutane

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