Explore topic-wise MCQs in Organic Chemistry.

This section includes 6 Mcqs, each offering curated multiple-choice questions to sharpen your Organic Chemistry knowledge and support exam preparation. Choose a topic below to get started.

1.

Which of the following structural formula that is consistent with the following observations: An acid, C7H4O2Cl2 shows a UV maximum of 242 nm?

A. <a href="https://www.sanfoundry.com/wp-content/uploads/2018/08/organic-chemistry-questions-answers-uv-visible-spectroscopy-q10a.png"><img alt="Formula with observations C7H4O2Cl2 shows UV maximum of 242 nm - option a" class="alignnone size-full wp-image-224175" height="126" src="https://www.sanfoundry.com/wp-content/uploads/2018/08/organic-chemistry-questions-answers-uv-visible-spectroscopy-q10a.png" width="79"/></a>
B. <a href="https://www.sanfoundry.com/wp-content/uploads/2018/08/organic-chemistry-questions-answers-uv-visible-spectroscopy-q10b.png"><img alt="Formula with observations C7H4O2Cl2 shows UV maximum of 242 nm - option b" class="alignnone size-full wp-image-224176" height="136" src="https://www.sanfoundry.com/wp-content/uploads/2018/08/organic-chemistry-questions-answers-uv-visible-spectroscopy-q10b.png" width="88"/></a>
C. <a href="https://www.sanfoundry.com/wp-content/uploads/2018/08/organic-chemistry-questions-answers-uv-visible-spectroscopy-q10c.png"><img alt="Formula with observations C7H4O2Cl2 shows UV maximum of 242 nm - option c" class="alignnone size-full wp-image-224177" height="120" src="https://www.sanfoundry.com/wp-content/uploads/2018/08/organic-chemistry-questions-answers-uv-visible-spectroscopy-q10c.png" width="80"/></a>
D. <a href="https://www.sanfoundry.com/wp-content/uploads/2018/08/organic-chemistry-questions-answers-uv-visible-spectroscopy-q10d.png"><img alt="Formula with observations C7H4O2Cl2 shows UV maximum of 242 nm - option d" class="alignnone size-full wp-image-224178" height="134" src="https://www.sanfoundry.com/wp-content/uploads/2018/08/organic-chemistry-questions-answers-uv-visible-spectroscopy-q10d.png" width="78"/></a>
Answer» D. <a href="https://www.sanfoundry.com/wp-content/uploads/2018/08/organic-chemistry-questions-answers-uv-visible-spectroscopy-q10d.png"><img alt="Formula with observations C7H4O2Cl2 shows UV maximum of 242 nm - option d" class="alignnone size-full wp-image-224178" height="134" src="https://www.sanfoundry.com/wp-content/uploads/2018/08/organic-chemistry-questions-answers-uv-visible-spectroscopy-q10d.png" width="78"/></a>
Discussion
2.

The ultraviolet spectrum of benzonitrile shows a secondary absorption band at 271 nm. If a solution of benzonitrile in water, with a concentration of 1 10-4 molar solution is examined at 271 nm, what will be the absorbance reading ( = 1000) and what will be the intensity ratio, IO/I, respectively?

A. 0.1, 1.26
B. 0.2, 2.26
C. 0.3, 3.26
D. 0.4, 4.26
Answer» B. 0.2, 2.26
Discussion
3.

The ultraviolet spectrum of benzonitrile shows a primary absorption band at 224 nm. If a solution of benzonitrile in water, with a concentration of 1x 10-4 molar, is examined at a wavelength of 224 nm, the absorbance is determined to be 1.30. The cell length is 1 cm. What is the molar absorptivity of this absorption band?

A. 2.3 x 10<sup>4</sup>
B. 3.3 x 10<sup>4</sup>
C. 1.3 x 10<sup>4</sup>
D. 4.3 x 10<sup>4</sup>
Answer» D. 4.3 x 10<sup>4</sup>
Discussion
4.

What is the correct order of max for n > * transition for the following three compounds?

A. RCOOH &gt; RCOOR &gt; RCONH<sub>2</sub>
B. RCOOH = RCOOR = RCONH<sub>2</sub>
C. RCOOH = RCOOR &lt; RCONH<sub>2</sub>
D. RCOOH = RCOOR &gt; RCONH<sub>2</sub>
Answer» C. RCOOH = RCOOR &lt; RCONH<sub>2</sub>
Discussion
5.

What is the correct order of max for n > * transition?

A. R-OH &gt; R-NH<sub>2</sub> &gt; R-SH
B. R-OH &lt; R-NH<sub>2</sub> &lt; R-SH
C. R-OH &gt; R-SH &gt; R-NH<sub>2</sub>
D. R-OH &lt; R-SH &lt; R-NH<sub>2</sub>
Answer» C. R-OH &gt; R-SH &gt; R-NH<sub>2</sub>
Discussion
6.

Which of the following comparison is correct for solvent shift on the n > * transition of acetone?

A. H<sub>2</sub>0 = CH<sub>3</sub>0H = C<sub>2</sub>H<sub>5</sub>0H = CHC1<sub>3</sub> = C<sub>6</sub>H<sub>14</sub>
B. H<sub>2</sub>0 &gt; CH<sub>3</sub>0H &gt; C<sub>2</sub>H<sub>5</sub>0H &gt; CHC1<sub>3</sub> &gt; C<sub>6</sub>H<sub>14</sub>
C. H<sub>2</sub>0 &lt; CH<sub>3</sub>0H &lt; C<sub>2</sub>H<sub>5</sub>0H &lt; CHC1<sub>3</sub> &lt; C<sub>6</sub>H<sub>14</sub>
D. H<sub>2</sub>0 &gt; CH<sub>3</sub>0H &lt; C<sub>2</sub>H<sub>5</sub>0H &lt; CHC1<sub>3</sub> &lt; C<sub>6</sub>H<sub>14</sub>
Answer» D. H<sub>2</sub>0 &gt; CH<sub>3</sub>0H &lt; C<sub>2</sub>H<sub>5</sub>0H &lt; CHC1<sub>3</sub> &lt; C<sub>6</sub>H<sub>14</sub>
Discussion