Question

quiz #4
tdsb.elearningontario.ca
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quiz information
question 5 (1 point)
which of the following compounds can exist as geometric isomers?
ch₂cl₂,
i
ch₂cl—ch₂cl,
ii
cl,
iii
ch₂cl—ch₂br
iv
i and iv
ii and iii
ii and iv
i, ii, iv
iv only
question 6 (1 point)
which of the following is a secondary alcohol?
oh
h₃c—c—ch₃
ch₃
h₃c—o—ch₃
oh
h₃c—c—ch₃
h
ch₃oh
ch₃ch₂oh

Explanation:

• Question 5: Geometric isomerism (cis - trans isomerism) occurs in alkenes (where there is restricted rotation around a double bond) or in cyclic compounds (where ring - strain restricts rotation).
• Compound I (\(CH_2Cl_2\)): It has a tetrahedral geometry (carbon is \(sp^3\) hybridized). There is free rotation around the \(C - Cl\) and \(C - H\) bonds (since it is a non - cyclic, \(sp^3\) hybridized carbon compound), so no geometric isomerism.
• Compound II (\(CH_2Cl - CH_2Cl\)): Both carbons are \(sp^3\) hybridized. There is free rotation around the \(C - C\) single bond (due to \(sp^3\) hybridization), so no geometric isomerism.
• Compound III (cyclic compound with a \(Cl\) substituent): In a cyclic compound (assuming it is a cycloalkane - like structure with \(sp^3\) hybridized carbons in the ring), there is no double bond. The ring itself has some restricted rotation, but for geometric isomerism in cyclic compounds, we need at least two non - identical substituents on different ring carbons. Here, the structure is not clear enough to have geometric isomerism (if it is a simple monosubstituted cycloalkane, no geometric isomerism).
• Compound IV (\(CH_2Cl - CH_2Br\)): If we assume it is \(ClCH = CHBr\) (maybe a mis - drawing, but if it is an alkene), there is restricted rotation around the double bond. Since the two substituents (\(Cl\) and \(Br\)) on each carbon of the double bond are different, it can show cis - trans (geometric) isomerism.
• Question 6: A secondary alcohol is an alcohol where the carbon atom bearing the \(-OH\) group is bonded to two other carbon atoms.
• For \(H_3C - \underset{CH_3}{\underset{|}{C}}-OH\): The carbon with the \(-OH\) group is bonded to three carbon atoms (it is a tertiary alcohol).
• For \(H_3C - O - CH_3\): This is an ether (\(C - O - C\) functional group), not an alcohol.
• For \(H_3C-\underset{H}{\underset{|}{C}}-OH - CH_3\): The carbon with the \(-OH\) group is bonded to two other carbon atoms (a methyl group and an ethyl group, for example, if we consider the full structure), so it is a secondary alcohol.
• For \(CH_3OH\): The carbon with the \(-OH\) group is bonded to one other carbon atom (it is a primary alcohol).
• For \(CH_3CH_2OH\): The carbon with the \(-OH\) group is bonded to one other carbon atom (it is a primary alcohol).

Answer:

• Question 5: IV only
• Question 6: \(H_3C-\underset{H}{\underset{|}{C}}-OH - CH_3\) (the third option in question 6)