How do i open an S3 link to share to Instagram?

I'm using React Native to create an iOS app. In the code below, I can easily access items in my gallery to open Instagram and share as a Story or on the Feed. However i would like to share a uri to an S3 video as follows:

import React from 'react'
import { Share, Linking ....} from 'react-native'

//Click button and run function
testShareInsta = async () => {
     const uri = "https://s3-dev.s3-us-west-1.amazonaws.com/video/myvideo";
     let encodedURL = encodeURIComponent(uri);  
     let instagramURL = 'instagram://library?AssetPath=${encodedURL}';
     return Linking.openURL(instagramURL);
  }

What's wrong here?