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<h2 class="hd hd-2 unit-title">Reference Summary</h2>
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<h3> <b> Polarity </b> </h3>
<p> Polarity is dependent on <br /> - Polar bonds <br /> - Geometry of the molecule </p>
<br />
<h3> <b> Dipoles </b> </h3>
<p> An unequal electron distribution is called a dipole. <br />Dipoles are often shown by a T-arrow or a δ. The dipoles are shown on the figure in red. The overall dipoles (which are a vector sum of the individual dipoles) are shown in blue.</p>
<img src='/assets/courseware/v1/184e6057f96249749893be9a7b8a4a5b/asset-v1:OCW+Pre-7.01+1T2020+type@asset+block/b00.png' alt="This image shoes two molecules, carbon dioxide C(=O)=O and water [H]O[H]. The dipoles and overal dipole of each molecule are drawn. For carbon dioxide, the C is labelled with a delta plus symbol to indicate a partial positive charge and the O atoms are labelled with delta minus symbols to indicate partial negative charges. There are two arrows shown to represent the dipoles. The non-arrowhead end of the arrow is crossed, similar to a plus sign. Each arrow starts at the C and points towards one of the O atoms. There is no overall dipole of carbon dioxide because the dipoles are equal in magnitude and opposite in direction. For water, the O is labelled with a delta minus symbol, and the two H atoms are labelled with delta positive symbols. Similar to carbon dixoide, water has two arrows with crossed tails. Each arrow starts at one of the H atoms and points towards the O atom. The H atoms are shown as bent downwards on the left and right sides of the O atom. Therefore, overall dipole is in the up direction.">
<br /><br />
<h3> For <b>CO<sub>2</sub></b>... <br /></h3> <p>The C=O bond is POLAR since O is more electronegative than C (attracts electrons more), BUT the vector sum of the dipoles is equal to zero, so CO<sub>2</sub> is NON-POLAR overall.
<h3> For <b>H<sub>2</sub>O</b>... <br /></h3> <p>The O-H bond is POLAR since O is more electronegative than H. The vector sum of dipoles is NOT equal to zero, so H<sub>2</sub>O is POLAR overall. </p>
<br />
<h3> <b> Example: Propane v. Propanol </b> </h3>
<img src='/assets/courseware/v1/d4d3fd8637291c7e66d0623321721c75/asset-v1:OCW+Pre-7.01+1T2020+type@asset+block/b3.png' alt="This image is of propane: CCC">
<p> This is <b>propane</b>. <br /> Chemical formula: C<sub>3</sub>H<sub>8</sub> <br /> Bonds: NON-POLAR, since C and H share electrons equally <br /> Molecule: NON-POLAR </p>
<br />
<img src='/assets/courseware/v1/ee3c0cdc87424246c979049c10867e7a/asset-v1:OCW+Pre-7.01+1T2020+type@asset+block/b4.png' alt="This image is of propanol: CCCO. The O is labelled with a delta minus to indicate that it has a partial negative charge symbol. The C and H that are attached to this O atom are each labelled with a delta plus symbol to indicate that they each have partial positive charges.">
<p> This is <b>propanol</b>. <br /> Chemical formula: C<sub>3</sub>H<sub>7</sub>OH or C<sub>3</sub>H<sub>8</sub>O <br /> Bonds: C-O and O-H bonds are POLAR, since O is more electronegative (attracts electrons more) than C or H <br /> Molecule: POLAR, the dipole vectors do not cancel out.
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<h2 class="hd hd-2 unit-title">Exercise 1 (Butane v. Butanol)</h2>
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<p> Let us now examine the molecules <b>butane</b>... </p>
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<p> ...and<b> butanol</b>.</p>
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Butane: Chemical Formula
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<p>Give the chemical formula of <b>butane</b>, shown above. (Enter answers as an integer.)</p>
<h3> How many C's? </h3>
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<h3> How many O's? </h3>
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Butane: Polarity
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<p>Does <b>butane</b> have any dipoles?</p>
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<p>Is <b>butane</b> POLAR or NON-POLAR?</p>
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Butanol: Chemical Formula
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<p>Give the chemical formula of <b>butanol</b>, also shown above. (Enter answers as an integer.)</p>
<p> How many C's? </p>
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<p> How many H's? </p>
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<p> How many O's? </p>
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Butanol: Polarity
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<p>Does <b>butanol</b> have any dipoles?</p>
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<p>Is <b>butanol</b> POLAR or NON-POLAR?</p>
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Butanol: Polar Group
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<p>Clicking on the image will place the circle on the image at the cursor location. <br/>Click on the image of butanol below to place the circle over butanol's polar group <em>(the one with unequal charge distribution)</em>. </p>
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Butanol: Dipoles
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<p>Describe the dipoles (&#948;) and each of their charges for the butanol molecule below. To do so, use the dropdown options to indicate the dipole (&#948;+ and &#948;-) for the corresponding region of the dipoles.</p>
<p><img src="/assets/courseware/v1/6b41eb5784947412dbbcd054a6383487/asset-v1:OCW+Pre-7.01+1T2020+type@asset+block/b6.png" width="400" style="float:center;" alt="Butanol."/> </p>
<ul> <li>the C adjacent to the -OH group:<div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div class="inputtype option-input inline">
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<option value="option_80ffa00ea42c4b0f9817ca7f503dc3aa_2_1_dummy_default">Select an option</option>
<option value="&#948;+"> &#948;+</option>
<option value="&#948;-"> &#948;-</option>
</select>
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<li>the O in the -OH group:<div class="inline" tabindex="-1" aria-label="Question 2" role="group"><div class="inputtype option-input inline">
<select name="input_80ffa00ea42c4b0f9817ca7f503dc3aa_3_1" id="input_80ffa00ea42c4b0f9817ca7f503dc3aa_3_1" aria-describedby="status_80ffa00ea42c4b0f9817ca7f503dc3aa_3_1">
<option value="option_80ffa00ea42c4b0f9817ca7f503dc3aa_3_1_dummy_default">Select an option</option>
<option value="&#948;+"> &#948;+</option>
<option value="&#948;-"> &#948;-</option>
</select>
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<li>the H in the -OH group:<div class="inline" tabindex="-1" aria-label="Question 3" role="group"><div class="inputtype option-input inline">
<select name="input_80ffa00ea42c4b0f9817ca7f503dc3aa_4_1" id="input_80ffa00ea42c4b0f9817ca7f503dc3aa_4_1" aria-describedby="status_80ffa00ea42c4b0f9817ca7f503dc3aa_4_1">
<option value="option_80ffa00ea42c4b0f9817ca7f503dc3aa_4_1_dummy_default">Select an option</option>
<option value="&#948;+"> &#948;+</option>
<option value="&#948;-"> &#948;-</option>
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<h2 class="hd hd-2 unit-title">Exercise 2 (Benzoic Acid)</h2>
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<p> Now let us examine <b> benzoic acid</b>.</p>
<img src='/assets/courseware/v1/7b82fda32549e9a4975dda749a99b24e/asset-v1:OCW+Pre-7.01+1T2020+type@asset+block/b7.png' alt="This image is of benzoic acid: C1=CC=C(C=C1)C(=O)O">
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Benzoic Acid: Bond Polarity
</h3>
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<option value="Yes"> Yes</option>
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Benzoic Acid: Molecule Polarity
</h3>
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<p>Would you predict that benzoic acid is a polar molecule? </p>
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<h2 class="hd hd-2 unit-title">Exercise 3 (Lysine)</h2>
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<p>Let us look at one more: <b> Lysine</b>.</p>
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Lysine: Molecule Polarity
</h3>
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<p>Would you predict that lysine is a polar molecule? </p>
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