A 30.0 mL sample of 0.200 M Ba(OH)2 is titrated with 0.100 M HCl.
1 answer:
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Answer:
See explanation
Explanation:
In this case, we have to check two variables:
1) The leaving group
2) The carbon bonded to the leaving group.
Let's check one by one:
<u>2-chloro-3-methylbutane</u>
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In this molecule, the leaving group is "Cl", the carbon bonded to the leaving group has two neighbors. Therefore, we have a <u>secondary substrate.</u>
<u>1-phenylpropan-1-ol</u>
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In this molecule, the leaving group is "OH", the carbon bonded to the hydroxyl group has two neighbors also. So, we have a <u>secondary substrate.</u>
<u>(E)-pent-3-en-2-yl 4-methylbenzenesulfonate</u>
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In this case, the leaving group is "OTs" (Tosylate), the carbon bonded to the tosylate group has as a neighbor a double bond. Therefore, we have an <u>allylic substrate.</u>
<u>3a-bromooctahydro-1H-indene</u>
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In this molecule, the leaving group is "Br", the carbon bonded to the bromine has three neighbors. So, we have a <u>tertiary substrate.</u>
<u>1-iodo-3-methylbutane</u>
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In this molecule, the leaving group is "I", the carbon bonded to the iodide has only one neighbor. So, we have a <u>primary substrate.</u>
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See figure 1
I hope it helps!
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