-Intermediate A is likely to have a lower molecular weight than the starting mat
ID: 560625 • Letter: #
Question
-Intermediate A is likely to have a lower molecular weight than the starting material. This can be predicted from the higher melting point.
-Intermediate A is likely to be water soluble due to its new bonds to Br.
-Intermediate A is likely to be much heavier (molecular weight) than E-stilbene due to the addition of two Br atoms.
-Intermediate A is a very strong base due to the two bromine atoms in its structure.
*What is the role of potassium hydroxide?
-strong base to perform the elimination reaction
-heterogeneous catalyst
-homogeneous catalyst
-good leaving group
-Intermediate A is likely to have a lower molecular weight than the starting material. This can be predicted from the higher melting point.
-Intermediate A is likely to be water soluble due to its new bonds to Br.
-Intermediate A is likely to be much heavier (molecular weight) than E-stilbene due to the addition of two Br atoms.
-Intermediate A is a very strong base due to the two bromine atoms in its structure.
*What is the role of potassium hydroxide?
-strong base to perform the elimination reaction
-heterogeneous catalyst
-homogeneous catalyst
-good leaving group
pyridinium perbromide MW 320 g/mole H'Br3 Addition: CH3CO2H mp 236-237 °C E-stilbene MW - 180 g/mole mp 123-125 Elimination: triethylene glycol diphenylacetylene MW- 178 g/mole mp=61Explanation / Answer
Since two bromine atoms are added to stilbene the molecular weight should increase hence the correct answer is:
"Intermediate A is likely to be much heavier (molecular weight) than E-stilbene due to the addition of two Br atoms."
*In the second reaction Elimination of Br takes place in the presence of KOH which is a strong base, Hence correct answer is
"strong base to perform the elimination reaction"
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.